What is the derivative of 2/3 (-2e^{-2x}+e^x) ?

The derivative of 2/3 (-2e^{-2x}+e^x) is 2/3 (4e^{-2x}+e^x)

What is the first derivative of 2/3 (-2e^{-2x}+e^x) ?

The first derivative of 2/3 (-2e^{-2x}+e^x) is 2/3 (4e^{-2x}+e^x)

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\frac{d}{d x} (\frac{2}{3} (- 2 e^{- 2 x} + e^{x}))

\frac{2}{3} (4 e^{- 2 x} + e^{x})

Solution steps

\frac{d}{d x} (\frac{2}{3} (- 2 e^{- 2 x} + e^{x}))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{2}{3} \frac{d}{d x} (- 2 e^{- 2 x} + e^{x})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{2}{3} (- \frac{d}{d x} (2 e^{- 2 x}) + \frac{d}{d x} (e^{x}))

\frac{d}{d x} (2 e^{- 2 x}) = - 4 e^{- 2 x}

\frac{d}{d x} (e^{x}) = e^{x}

= \frac{2}{3} (- (- 4 e^{- 2 x}) + e^{x})

Simplify

= \frac{2}{3} (4 e^{- 2 x} + e^{x})

\int (sin (y)) d y

\int cos^{5} (x t a) n^{3} x d x

\frac{x ^{4} + 1}{x ^{2} + 1}

y = x, y = 1 - x, y = - \frac{x}{2}

What is the derivative of 2/3 (-2e^{-2x}+e^x) ?
The derivative of 2/3 (-2e^{-2x}+e^x) is 2/3 (4e^{-2x}+e^x)
What is the first derivative of 2/3 (-2e^{-2x}+e^x) ?
The first derivative of 2/3 (-2e^{-2x}+e^x) is 2/3 (4e^{-2x}+e^x)